Download Homework Section 2-1

Chapter Two
Deductive Reasoning
Objectives
A. Use the terms defined in the chapter
correctly.
B. Properly use and interpret the
symbols for the terms and concepts
in this chapter.
C. Appropriately apply the properties
and theorems in this chapter.
D. Correctly interpret the information contained in a
conditional.
E. Understand and properly use algebraic properties
F. Properly prove theorems relating to lines and angles.
Section 2-1
If-Then Statements; Converses
Homework Page 35:
1-30
Objectives
A. Identify and properly use
conditional statements.
B. Identify the hypothesis and
conclusion of conditionals.
C. Identify and state the converse of
a conditional.
D. Provide correct counterexamples to prove statements false.
E. Identify and use biconditional statements.
Conditional Statements
• We use conditional statements in our everyday language,
as well as in our mathematical language.
• The common form of a conditional statement, or
conditional, is:
– If hypothesis, then conclusion.
Conditional Statements
• Hypothesis
– According to Merriam-Webster dictionary  a
tentative assumption made in order to draw out and test
its logical or empirical consequences
– More simply  a set of pre-conditions from which we
attempt to reach a conclusion.
– Or  the information which must be known in order to
apply the conclusion to a problem.
– In geometry, it is common for the hypothesis to
describe a diagram, or to be part of a diagram.
• Conclusion  Information which can be added to a
problem when the criteria of the hypothesis have been met.
Conditional Statements
• Conditional statements can be TRUE or FALSE:
– If they are considered TRUE, they must be TRUE in
ALL cases.
– If there is a SINGLE case where the statement is false,
then the ENTIRE conditional is considered FALSE!
• Examples:
– If I do not eat, then I will eventually starve.
– If I live in Bexley, then I live in Ohio.
– If I add 3 to 4, then I will have 7.
– If you cheat on homework, then you won’t do well in
this class.
– If you want the freedom of an adult, then you must
accept adult responsibilities.
Other Forms of Conditional Statements
• Thanks to the English language, you have several other
ways of expressing a conditional statement:
– IF hypothesis, THEN conclusion.
– hypothesis IMPLIES conclusion.
– hypothesis ONLY IF conclusion.
– conclusion IF hypothesis.
Equivalent Conditionals: Examples
•
•
•
•
If you live in Ohio, then you live in the United States.
You live in Ohio implies that you live in the United States.
You live in Ohio only if you live in the United States.
You live in the United States if you live in Ohio.
4
“Say it ain’t so!”
• A major outcome of your work in this class will be your ability to
prove or disprove conditionals.
• Remember, a conditional is always true or it is false, there is no
“sometimes this, sometimes that”.
• To prove a conditional or theorem to be true usually takes a number of
steps.
– The proof MUST show that the statement to be true for ALL cases.
• To prove something is false we need ONLY ONE example where the
hypothesis contradicts the conclusion.
– Such an example is known as a counterexample:
• A counterexample proves a conditional false by agreeing with
the hypothesis but disagreeing with the conclusion.
Counterexamples
• If it is a week night, then you have geometry homework.
– Counterexample  December 25th may be a week
night this year, but you don’t have geometry homework
on Christmas.
• This statement agrees with the hypothesis, but disagrees
with the conclusion.
• Since we have found one counterexample for the
conditional we say the conditional is false.
• If we find at least one counterexample that proves the
conditional false, it makes no difference how many
examples we can find where it is true.
• Because the conditional was false once, it may be false
again. So the conditional has no value in predicting the
future or judging the present.
4
Converse of a Conditional
 converse: The converse of a conditional is another if-then
statement formed by interchanging the hypothesis and the
conclusion of a given statement.
Conditional statement  If p then q
Converse of above statement  If q then p
 Example:
Conditional
If tomorrow is Saturday, then today is Friday.
Converse
If today is Friday, then tomorrow is Saturday.
Converses of Conditionals
• NOTE!
Just because the conditional statement is true
does NOT make the converse of the statement true!
– Likewise, just because the converse of a statement is
true does not make the conditional true.
• Example:
– Conditional  If I have 2 dimes and a nickel, then I
have 25 cents.
– Converse  If I have 25 cents, I have 2 dimes and a
nickel.
• Remember, you need only ONE counterexample to prove a
statement false.
Biconditionals
• For a statement to be biconditional, both the original
conditional (statement) and its converse must be true.
• One sign that you have a biconditional statement is the key
phrase “if and only if” to connect the parts of the
statement.
• In a biconditional, the order of the phrases can be switched
without changing the meaning.
Biconditional Example
• Conditional  If I draw a right angle, then I draw a 90
degree angle.
• Converse  If I draw a 90 degree angle, then I draw a
right angle.
• Biconditional  I draw a right angle if and only if I draw a
90 degree angle.
Definitions Written as Biconditionals
• An angle is acute if and only if it measures between 0 and
90.
– An angle measures between 0 and 90 if and only if
the angle is acute.
• A ray bisects an angle if and only if it divides the angle
into two congruent adjacent angles.
– A ray divides the angle into two congruent adjacent
angles if and only if the ray bisects an angle.
• Points are collinear if and only if the points lie on one line.
– Points lie on one line if and only if the points are
collinear.
3
Recommendations
• When it is possible, I HIGHLY recommend that you
reword any definitions, postulates, and theorems in their
biconditional form.
• In all other cases, I HIGHLY recommend you reword any
definitions, postulates, and theorems in their IF/THEN
form.
Section 2-1 Sample Problems
Write the hypothesis and the conclusion of each
conditional.
1. If 3x - 7 = 32, then x = 13
If 3x - 7 = 32, then x = 13
3. I’ll try if you will.
I’ll try if you will.
5. a + b = a implies b = 0.
a + b = a implies b = 0.
Section 2-1 Sample Problems
Rewrite each pair of conditionals as a biconditional.
7. If B is between A and C, then AB + BC = AC.
If AB + BC = AC, then B is between A and C.
B is between A and C if and only if AB + BC = AC.
AB + BC = AC if and only if B is between A and C.
Write each biconditional as two conditionals that are
converses of each other.
9. Points are collinear if and only if they all lie on one line.
If points are collinear, then they all lie on one line.
If points all lie on one line, then they are collinear.
Sample Problems Section 2-1
Provide a counterexample to show that each statement is
false. You may use words or diagrams.
11. If ab < 0, then a < 0.
Let a = 2 and b = -3. Therefore, ab = -6.
In this case, ab < 0 (agrees with hypothesis), but a > 0
(disagrees with conclusion).


13. If point G is on AB, then G is on BA .
15. If a four sided figure has four right angles, then it has
four congruent sides.
Sample Problems Section 2-1
Tell whether each statement is true or false. Then write
the converse and tell whether it is true or false.
17. If x = - 6, then x  = 6.
True
If | x | = 6, then x = -6.
False, x = 6.
19. If b > 4, then 5b > 20.
21. If Pam lives in Chicago, then she lives in Illinois.
23. a2 > 9 if a > 3.
Sample Problems Section 2-1
25. n > 5 only if n > 7.
27. If points D, E and F are collinear, then DE + EF = DF.
29. Write a definition of congruent angles as a biconditional.
Section 2-2
Properties from Algebra
Homework Pages 41-42:
1-14
Objectives
A. Properly use and describe
algebraic properties.
B. Relate the algebraic properties to
geometric properties.
C. Properly apply geometric
properties.
Algebraic Properties of Equality: Transformations
• Addition: You may add the same value to both sides of an
equation.
– If a = b, then a + c = b + c.
• Subtraction (add a negative): You may subtract the same
value from both sides of an equation.
– If a = b, then a - c = b - c.
• Multiplication: You may multiply the same value to both
side of an equation.
– If a = b, then a * c = b * c.
Algebraic Properties of Equality: Transformations
• Division (multiply by a reciprocal): You may divide the
same value into both sides of an equation.
– If a = b, then a / c = b / c.
– HOWEVER: c cannot be zero!
• Distribution: You may multiply a factor next to a grouping
symbol to every term inside the grouping symbol.
– If a (b + c + d) = e, then ab + ac + ad = e.
• Substitution: Left and right sides of an equation are
interchangeable. Either statement may be used in another
equation.
– If a + b = c AND d – e = c, then:
• a+b=d–e
Algebraic Properties of Equality: Transformations
• Reflexive: A value must equal itself.
a=a
• Symmetric: The left and right sides of an equation can be
switched.
– If a = b, then b = a.
• Transitive: Any two values in a chain of equality are
equal.
– If a = b AND If b = c. then a = c.
But what about GEOMETRIC properties?
• Remember, we cannot talk about two FIGURES being
EQUAL!
– Two geometric figures can be CONGRUENT.
– So certain properties, such as the Addition Property of
Equality, cannot be applied to figures.
• However, SOME algebraic properties can be applied to
figures.
– Lengths of line segments and measures of angles are
real numbers.
• Therefore, we can apply algebraic properties of
equalities (such as the Addition Property) to these
real numbers.
Properties of Congruence
•Reflexive: Any object must be
congruent (same size and shape) to
itself.
•Symmetric: The objects on the
left and right sides of a congruence
statement may be switched.
DE  DE
D  D
If DE  FG then FG  DE .
If D  E then E  D.
•Transitive: Any two objects in a
chain of congruence statements are
congruent (same size and shape).
If DE  FG and FG  JK then DE  JK .
If D  E and E  F then D  F .
So, what do we do with these properties?
• We use these algebraic and geometric properties to
prove statements.
• For example:
If 3X + 5 = 17, then X = 4.
Given.
3X + 5 (- 5) = 17 (- 5)
3X = 12
Subtraction Property of
Equality.
3 X 12 X

3
3
4 X
Division Property of
Equality.
X=4
Symmetric Property of
Equality.
Sample Problems Section 2-2
Justify each step. (Give the reason for each step.)
1.
4x – 5 = -2
Given.
4x = 3
Addition Property of Equality
3
x
4
Division Property of Equality
Sample Problems Section 2-2
Justify each step. (Give the reason for each step.)
3.
z7
 11
3
Given.
z  7  33
Multiplication Property of Equality
z = -40
Addition Property of Equality
Sample Problems Section 2-2
Justify each step. (Give the reason for each step.)
5.
2
b  8  2b
3
2b  3(8  2b)
2b  24  6b
8b  24
b3
Sample Problems Section 2-2
Given:  AOD as shown
Prove: m  AOD = m  1 + m  2 + m  3 A 
B
1
2
C
3
O
Statements
0.  AOD as shown
1. m  AOD = m  AOC + m  3
2. m  AOC = m  1 + m  2
3. m  AOD = m  1 + m  2 + m  3

D

Reasons
0.
1.
2.
3.
Given.
Angle Addition Postulate
Angle Addition Postulate
Substitution Property of Equality
Sample Problems Section 2-2
Given: DW = ON
Prove: DO = WN
Statements
1. DW = ON
2. DW = DO + OW
ON = ___ + ___
3. ________________
4. OW = OW
5. ________________
D
O
W
N
Reasons
1.
2.
3. Substitution
4.
5.
Sample Problems Section 2-2
S
11. Given: m  1 = m  2;
m3=m4
Prove: m  SRT = m  STR
P
Q
Z
4
3
R
1. m1  m2; m3  m4
2. mSRT  m1  m3;
mSTR  m2  m4
3. mSRT  m2  m3
4. mSRT  m2  m4
5. mSRT  mSTR
1
1. Given
2. Angle Addition Postulate
3. Substitution Property of =
4. Substitution Property of =
5. Substitution Property of =
2
T
Sample Problems Section 2-2
13. Given: RQ = TP
ZQ = ZP
Prove: RZ = TZ
S
Q
P
Z
4
3
1
R
2
T
Section 2-3
Proving Theorems
Homework Page 46:
1-16
Objectives
A. Use the Midpoint Theorem and
the Angle Bisector Theorem
correctly.
B. Understand the valid reasons used
in proofs.
C. Apply valid reasons to prove
theorems and conditionals.
Deductive Reasoning
• Also known as direct proof
• Deductive reasoning is one logical process used to prove
conditionals true by building an argument based upon valid
reasoning.
Valid Reasons Used in Proofs
• The items you may use in a proof are:
– Information given in the hypothesis,
– Information given in diagrams,
– Accepted postulates,
– Algebraic and geometric properties,
– Definitions,
– Previously proven or accepted theorems, and
– Previously proven or accepted corollaries.
Theorem 2-1 (Midpoint Theorem)
If M is the midpoint of segment AB,
A
M
then AM = ½AB and MB = ½AB.
B
Proof Of Midpoint Theorem
If M is the midpoint of segment AB,
then AM = ½AB and MB = ½AB.
A
M
Now that I know M is the midpoint, what else can I say?
Statements
M is the midpoint of AB
Reasons
Given
AM  MB
Definition of Midpoint.
AM  MB
Definition of Congruence.
AM + MB = AB
Segment Addition Postulate.
AM + AM = AB
2AM = AB
Substitution Property.
Simple addition.
AM = ½ AB
Division Property of Equality
B
 Theorem 2-2 (Angle Bisector Theorem)
If ray BX is the bisector of ABC,
A
X
B
C
then mABX = ½mABC and mXBC = ½mABC.
Proof Of Angle Bisector Theorem
A
If ray BX is the bisector of ABC,
then m  ABX = ½ m  ABC and
m  XBC = ½ m  ABC.
X
B
Statements
Reasons
C
Sample Problems Section 2-3
Name the definition, postulate or theorem that justifies
the statement about the diagram.
1. If D is the midpoint of BC , then BD  DC

3. If AD bisects  BAC, then  1   2.
5. If BD  DC ,
A
1 2
then D is the midpoint of BC
7. m  1 + m  2 = m  BAC
1. Definition of midpoint.
3. Definition of angle bisector.
5. Definition of midpoint.
7. Angle Addition Postulate
B
3 4
D
C
Sample Problems Section 2-3
Write the number that is paired
C
with the bisector of
 CDE.
9.
9.
80  40
 60
2
11.
0
180
E
D
11.
C
0
180
E
D
122  18
 70
2
Sample Problems Section 2-3
13. The coordinates of points L and X are 16 and 40,
respectively. N is the midpoint of LX, and Y is the midpoint
of LN. Sketch a diagram and find:
a. LN b. the coordinate of N c. LY d. the coordinate of Y
16
40
X
N
Y
LN = Length of the segment LN = | 16 – 40 | = | -24 | = 24
N = Midpoint of the segment LX = (40 + 16) / 2 = 28
LY = Length of the segment LY = ½ LN
LN = ½ LX = ½ (24) = 12
LY = ½ LN = ½ (12) = 6
Y = Midpoint of the segment LN = (16 + 28) / 2 = 22
L
Sample Problems Section 2-3
15.a. Suppose M and N are the midpoints of LK and GH,
respectively. What segments are congruent?
b. What additional information about the picture would
enable you to deduce that LM = NH.
G
LM = NH
LM = GN
N
H
MK = NH
LK = GH
Sample Problems Section 2-3
What can you deduce from the given information?
17. Given: AE = DE;
A
CE = BE
B
E
D
AE = BE?
DE = EC?
AC = DB?
C
Section 2-4
Special Pairs of Angles
Homework Pages 52-54:
1-33
Objectives
A. Use the terms complementary,
supplementary, and vertical angles
correctly.
B. Apply these terms to proofs.
C. Use the Vertical Angle Theorem
correctly.
Definitions
complementary angles: A pair of coplanar angles, called
complements, whose measurements add up to be 90°.
supplementary angles: A pair of coplanar angles whose
measurements add up to be 180°. A supplement of an
angle is another angle that when added to the first makes
180°.
vertical angles: A pair of coplanar angles such that the sides
of one angle are opposite rays to the sides of the other
angle.
Complementary Angles
3
Supplementary Angles
6
Vertical Angles
6
 Theorem 2-3
Vertical angles are congruent.
Proof Of Theorem 2-3 (Vertical Angles Theorem)
Given: Diagram
Prove: 1  2
3
1
Statements
2
Reasons
1. Diagram
1. Given
2. Angles 1 & 2 are vertical angles
2. Definition of vertical angles
3. m1  m3  180
m2  m3  180
4. m1  m3  m2  m3
5. m3  m3
3. Definition of straight angles
(Angle Addition Postulate)
4. Substitution Property of Equality
5. Reflexive Property of Equality
6. m1  m2
6. Subtraction Property of Equality
7. 1  2
7. Definition of congruence
Sample Problems Section 2-4
Find the measures of a complement and a supplement of  K.
1. m  K = 20
3. m  K = x
5. Two complementary angles are congruent. Find their measures.
1. Complimentary angles add to 90°.
Supplementary angles add to 180°.
Compliment + m  K = 90° Supplement + m  K = 180°
Compliment + 20° = 90°
Supplement + 20° = 180°
Compliment = 70°
Supplement = 160°
3. Compliment + m  K = 90° Supplement + m  K = 180°
Compliment + x° = 90°
Supplement + x° = 180°
Compliment = 90° - x°
Supplement = 180° - x°
5. x° + x° = 90°
2x° = 90°
x° = 45°
Sample Problems Section 2-4
In the diagram,  AFB is a right angle. Name the figures
described.
7. Another right angle.
9. Two congruent supplementary angles.
11. Two acute vertical angles.
7. AFD
A
E
9. AFD and AFB
11. EFD and BFC
How can you PROVE
these angles are acute and vertical?
F
B
C
D
Sample Problems Section 2-4

In the diagram, OT bisects  SOU, m  UOV = 35, and
m  YOW = 120. Find the measure of each angle.
13. m  ZOY
17. ½ (120°) = 60°
15. m  VOW
U
T
17. m  TOU
V
35°
S
W
O
15. 180° - 35° - 120° = 25°
Z
Y
X
13. 35°
120°
Sample Problems Section 2-4
Find the value of x.
19.
21.
64°
36°
70
19. (3x – 5)° = 70°
3x° = 75°
x° = 25°
4x°
21. 64° + 36° = 100° WHY?
4x° = 100° WHY?
x° = 25° WHY?
Sample Problems Section 2-4
23. Given:  2   3
Prove:  1   4
2
1
0.  2   3 and diagram
1.  1   2
2.  2   3
3.  3   4
4.
3
4
0.
1.
2.
3.
4. Transitive Property
Sample Problems Section 2-4
If  A and  B are supplementary, find the value of x,
m  A and m  B.
25. m  A = x + 16, m  B = 2x - 16
25. m  A + m  B = 180°
x° + 16° + 2x° - 16° = 180°
REASON?
REASON?
3x° = 180°
REASON?
x° = 60°
REASON?
m  A = 76° m  B = 104°
REASON?
Sample Problems Section 2-4
If  C and  D are complementary, find the value of y,
m  C and m  D.
27. m  C = y - 8, m  D = 3y + 2
Use the information given to write an equation and solve
the problem.
29. Find the measure of an angle that is half as large as its
complement.
31. A supplement of an angle is six times as large as a
complement of the angle. Find the measures of the angle,
its supplement and its complement.
Sample Problems Section 2-4
Find the values of x and y for each diagram.
33.
50°
(3x - y)°
x°
Section 2-5
Perpendicular Lines
Homework Pages 58-60:
1-25, 28
Objectives
A. Use the term perpendicular lines
correctly.
B. Apply the term perpendicular
lines to theorems.
C. Use the theorems associated with
perpendicular lines (Theorems
2-4, 2-5, and 2-6) correctly.
Definition of Perpendicular Lines
• Perpendicular lines are two lines that intersect to form
right angles.
– Notice how we are using previous terms (defined and
undefined) to build new terms.
Theorem 2-4
If two lines are perpendicular, then they form congruent
adjacent angles.
Theorem 2-5
If two lines form congruent adjacent angles, then the lines are
perpendicular.
Comparing Theorems 2-4 and 2-5
• How is Theorem 2-4 (If two lines are perpendicular, then
they form congruent adjacent angles) related to Theorem
2-5 (If two lines form congruent adjacent angles, then the
lines are perpendicular)?
– Are they the same?
– Could they be written as a single statement?
– What type of statement?
Theorem 2-6
If the exterior sides of two adjacent acute angles are
perpendicular, then the angles are complementary.
A
D
B
C
mABD + m DBC = 90°
Proof Of Theorem 2-6: If the exterior sides of two adjacent acute angles
are perpendicular, then the angles are complementary.
Given : OA  OC
Prove : AOB and BOC
are compliment ary.
A
O
Statements
1. OA  OC
2. AOC is a right angle.
3. mAOC  90.
B
C
Reasons
1. Given
2. Definition of  lines.
3. Definition of right angle.
4. mAOB  mBOC  mAOC 4. Angle addition postulate.
5. mAOB  mBOC  90
5. Substitution property of =
6. mAOB and mBOC complimentarys
6. Def . of complimentary s.
Sample Problems Section 2-5
1. In the diagram, UL  MJ and m  JUK = x°. Express in
terms of x the measures of the angles:
a.  LUK
b.  MUK
a. mLUK  90  x
b. mMUK  180  x
L
K
x°
M
U
J
Sample Problems Section 2-5
Name the definition or state the theorem that justifies the
statement about the diagram.
3. If  EBC is a right angle, then BE  AC.
5. If BE  AC, then  ABD and  DBE are complementary.
7. If BE  AC, then m  ABE = 90°.
D
E
F
A
B
C
3. Definition of perpendicular lines.
5. If the exterior sides of 2 adjacent acute angles are perpendicular,
then the angles are complementary.
7. Definition of perpendicular lines and right angles.
Sample Problems Section 2-5
In the diagram, BE  AC and BD  BF.
Find the value of x.
9. m  ABD = 2x - 15, m  DBE = x
11. m  ABD = 3x - 12, m  DBE = 2x + 2,
m  EBF = 2x + 8
D
A
E
F
B
C
9. m  ABD + m  DBE = 90° 11. m  ABD + m  DBE = 90°
2x - 15 + x = 90°
3x - 12 + 2x + 2 = 90°
3x - 15 = 90°
5x - 10 = 90°
3x = 105°
5x = 100°
x = 35°
x = 20°
Sample Problems Section 2-5
In the figure BF  AE, m  BOC = x, and m  GOH = y.
Express the measures of the angles in terms of x and y.
15.  COH
17.  DOE
B
C
15. m COH = 180° - y°
x°
D
17. m DOE = 90° - (x° + y°)
A
O
y°
H
G
F
E
Sample Problems Section 2-5
Can you conclude from the information given for each
exercise that XY  XZ?
19.  1 and  3 are complementary
21. m  1 = m  4
Y
23. m  1 = m  2 and m  3 = m  4
25.  1   4 and  2   3
1
2
19. No, m 1 = 40° m 2 = 10° m 3 = 50° X
21. No, what if m  1 = 70°?
3
4
23. Yes:
m  1 + m  2 + m  3 + m  4 = 180°
Z
2m  2 + 2m  3 = 180°
m  2 + m  3 = 90°
25. No, what if m  1 = 70°?
Sample Problems Section 2-5
What can you conclude from the information given.
27. Given: AD  AC; CE  AC; m  1 = m  4
B
D
E
1
4
2
A
3
C
Section 2-6
Planning a Proof
Homework Pages 63-64:
1-22
Objectives
A. Understand and apply theorems
2-7 (supplementary angles) and 28 (complementary angles)
correctly.
B. Apply the two-column deductive
proof method to prove statements,
theorems, and/or corollaries.
Theorem 2-7
If two angles are supplements of congruent angles (or the
same angle), then the two angles are congruent.
A
D
B
C
ABC & BCD are supplementary
ADC & BCD are supplementary
ABC & ADC are congruent
Theorem 2-8
If two angles are complements of congruent angles (or of the
same angle), then the two angles are congruent.
B
C
A
O
D
AOB & BOC are complementary
COD & BOC are complementary
AOC & COD are congruent
Producing a Proof
• All proofs you will produce will be two-column proofs,
regardless of the directions in the book.
– Therefore, if the book says produce a paragraph proof,
you will instead produce a two-column proof.
• All two-column proof should appear as a two-column table
• The left column will be the statements column.
• The right column will be the reasons column.
Statements
Reasons
Parts of a Two Column Proof
Statements:
Each statement should be numbered and supported in the
reason column.
The first statement is a list of the given information found
in the hypothesis of the conditional or from
information in a diagram.
The body of the proof consists of a logical series of
statements and reasons flowing from the givens and
from information that can be proved from the diagram.
The last statement must be what you were required to
prove, found in the conclusion of the conditional.
Parts of a Two Column Proof
Reasons:
Each reason should be numbered so that it matches the
statement it supports.
The first reason will be “given” provided that the first
statement was a list of the given information.
Only “given”, definitions, postulates, algebraic properties,
theorems and their corollaries are acceptable reasons
for the body of the proof.
The final reason will depend upon the logical structure of
the whole proof, but it still must come from the list
above (except that it cannot be “given” ).
Each Row in the Two-column Proof
• Remember! For each row in the proof:
– The hypothesis in the reason must be shown to be true in preceding
statements.
– The conclusion in the reason must match the statement.
• Writing all definitions, postulates, properties, theorems and corollaries
as standard if-then conditionals will make this a lot easier!
C
A
O
D
Statements
B
Reasons
1. Given lines AB and CD are
perpendicular and intersect at
point O.
1. Given
2. Angle BOC is a right angle.
2. If two lines are perpendicular
then they intersect at right angles.
Steps for Writing a Two Column Proof
Step 1: Identify the conditional you are required to prove.
Step 2: Draw, label, and annotate a diagram for the proof.
Step 3: List from the conditional, in terms of the figure, what is given.
– This is the first statement/reason in the proof.
Step 4: Determine from the conditional, in terms of the diagram, what is to
be proven.
– This will be the last statement in the proof.
Step 5: List from the diagram, in terms of the figure, what can be proven and
why.
Step 6: Select and arrange, in a logical order, those statements from steps 3,
4 & 5 that will allow you to move from the given information to the
statement to be proved.
N.B. If you get stuck writing the proof from the top down, then work up 22
from the conclusion by deciding what you had to know to make that
statement.
N
Example of a Two Column Proof: #21 page 64
Step 1: If AC  BC and 3 is complementary to 1, then 3  2.
Step 2:
Step 3: Given: AC  BC
3 is complementary to 1
C
1 2
Step 4: Prove: 3  2
3
A
D
B
Step 5: mACB = 90, Definition of perpendicular lines, right angles
m3 + m 1 = 90, def comp ‘s
m2 + m 1 = m ACB, Angle Addition Postulate
m2 + m 1 = 90, substitution property of equality
2 is complementary to 1, def comp ‘s
m3 + m 1 = m2 + m 1, transitive
m3 = m2, subtraction property of equality
3  2, Th. 2-8: Complementary Angles
P
Example of a Two Column Proof: #21 page 64
Step 5: mACB = 90, definition of right angles, definition of 
m3 + m 1 = 90, def comp ‘s
m2 + m 1 = m ACB, Angle Addition Postulate
m2 + m 1 = 90, substitution property of =
2 is complementary to 1, def comp ‘s
m3 + m 1 = m2 + m 1, transitive property of =
m3 = m2, subtraction property of equality
3  2, Th. 2-8: Complimentary Angles
Step 6:
Statements
1. AC  BC, 3 is complementary to 1
2. mACB = 90
3. m2 + m 1 = m ACB
4. m2 + m 1 = 90
5. 2 is complementary to 1
6. 3  2
Reasons
1. Given
2. def 
3. Angle Add. Post.
4. Substitution prop =
5. def comp ‘s
6. Th. 2-8
P
Sample Problems Section 2-6
Write the name or statement of the definition, postulate,
property, or theorem that justifies the statement about the
diagram.
1. AD + DB = AB
C
3.  2   6
1. If D is on the line segment AB,
then AD + DB = AB.
3. If vertical angles exist,
then the angles are congruent.
A
H
F
D 1 2
6
3
5 4
B
G
E
Sample Problems Section 2-6
Write the name or statement of the definition, postulate,
property, or theorem that justifies the statement about the
diagram.

5. If DF bisects  CDB, then  1   2.
C
7. If CD  AB, then m  CDB = 90°.
7. If two lines are perpendicular,
then they meet to form right angles.
If an angle is a right angle,
then it measures 90°.
A
H
5. If a ray bisects an angle,
then it cuts the angle into 2 congruent angles.
F
D 1 2
6
3
5 4
B
G
E
Sample Problems Section 2-6
Write the name or statement of the definition, postulate,
property, or theorem that justifies the statement about the
diagram.
9. If m  3 + m  4 = 90, then
 3 &  4 are complementary.
C
11. If AB  CE, then  ADC   ADE.
13. If  FDG is a right angle, then DF  DG.
D 1 2
9. If the sum of the measure of two
6
3
A
5 4
angles is 90°,
H
then the angles are complementary.
E
11. If two lines are perpendicular,
then they form congruent adjacent angles.
F
B
G
Sample Problems Section 2-6
15. Given:  1 &  5 are supplementary
 3 &  5 are supplementary
Prove:  1   3
1
1.  1 &  5 are supplementary
 3 &  5 are supplementary
2. m  1 + m  5 = 180
m  3 + m  5 = 180
3. m  1 + m  5 = m  3 + m  5
4. m  5 = m  5
5. m  1 = m  3 or  1   3
3
1.
2.
3.
4.
5.
5
Sample Problems Section 2-6
17. Given: PQ  QR, PS  SR,  1   4
Prove:  2   5
Q 3
1 2
R
P
4 5
S 6
1. PQ  QR, PS  SR,  1   4
2.  2 is comp. to  1
 5 is comp to  4
3.  2   5
1.
2.
3.
Sample Problems Section 2-6
19. Given:  2   3
Prove:  1   4
21. Given: AC  BC
 3 is comp. to  1
Prove:  3   2
1
2
3
4
C
12
3
A
D
B
Chapter Two
Deductive Reasoning
Review
Homework Pages 68-69:
1-12